如何将泡菜文件上传并保存到 django rest 框架？

Question

As it says in the title, I am trying to upload a model.pkl and save it in an API made in django. I manage to save the model.pkl correctly in the API, however the file is uploaded in a corrupt way, because I cannot read the pickle file.

Im open to any solutions that can make the pickle file be uploaded, stored and readed in the API.

Class to upload the file and save it

class FileUploadView(APIView):
    parser_classes = (FileUploadParser,)

    def put(self, request):
        file = request.data.get('file', None)
        file_name = f'path/{file.name}'

        path = default_storage.save(name=file_name, content=ContentFile(file.read()))
        tmp_file = os.path.join(settings.MEDIA_ROOT, path)

        if file is not None:
            return Response(f'File: {file.name} successfully uploaded and saved!', status=HTTP_200_OK)
        else:
            return Response(f'File not found!', status=HTTP_400_BAD_REQUEST)

Error when reading the file

def read_PickleModel(path_to_PickleModel):
    with open(path_to_PickleModel, 'rb') as f:
        pickle_model = pickle.load(f)
    return pickle_model

read_PickleModel(path_to_PickleModel)

Traceback:
DEBUG: An exception has ocurred: invalid load key, '-'.

Im new in this, please before voting down, tell me how do you think I can improve the question to get an accurate solution.

Postman

Answer 1

When using FileUploadParser the entire body of the request should be the file contents, don't send the file as form data.

To do this in Postman, select "binary" as the data type instead of "form-data" and select your file there

To set the filename you need to set a header "Content-Disposition" and pass attachment; filename=<your_filename.ext> attachment; filename=<your_filename.ext>